Master Degree in Mechanical Engineering Numerical Heat and Mass Transfer 16 Convective Heat Transfer Fausto Arpino f.arpino@unicas.it
Forced convection over a flat plate Assuming steady-state regime, two-dimensional flow, incompressible fluid with constant properties, the velocity and thermal fields are governed by the following equations: Mass conservation equation u v + = 0 x y Momentum conservation equation along x-direction u u p u u ρ u + v = Fx x y + µ + x x y Momentum conservation equation along y-direction v v p v v ρ u + v = Fy + µ + x y y x y Energy conservation equation T T T T cρ u + v = k + x y x y Dimensional form Mass conservation equation u* v* + = 0 x* y* Momentum conservation equation along x-direction u* u* p* 1 u* u* u* + v* = + + x* y* x* Re x* y* Momentum conservation equation along y-direction v * v* p* 1 v* v* u* + v* = + + x * y* y* Re x* y* Energy conservation equation T T T T u* + v* = + x* y* Re Pr x* y* * * 1 * * Dimensionless form (for forced convection) 2
Forced convection over a flat plate The scales and parameter used to derive the non dimensional equations are: x* = x L y* = y L p* = p ρu 2 T* = T T ΔT Re = u L ν u* = u u Pr = ν a v* = v u The forced convection problem has been deeply investigated by Prandtl, who concluded that velocity gradients are important only within the boundary layer, while can be neglected outside such zone. Prandtl based the analysis on the order of magnitude investigation of different terms in the governing equations, observing that the velocity and thermal boundary layers can be considered negligible if compared to the characteristic dimension of the body δ << L, δ t << L and that x * 1 O( u *) 1, O( x *) 1 O u * From mass conservation equation... O v * y * 1 3
Forced convection over a flat plate Since the order magnitude of y* is δ*, also the order of magnitude of v* must be δ *. O( y *) =O( δ *) where δ* = δ L O( v *) δ* << 1 velocity y-component is negligible with respect to its x-component This means that in the boundary layer the y-component of the velocity is much smaller than the its x- component, and only the x-momentum conservation equation can then be considered. Observing that all the terms in one equation must have the same order of magnitude, it can be derived from the momentum conservation equation in the x-direction that all the terms in the right hand side of the equation must present order of magnitude not larger than unity. O(...) 1 O(...) 1 u * u * x * + v * u * y * = p * x * + 1 Re 2 u * x * + 2 u * 2 y * 2 O v * u * y * 1 u * y * = 1 δ * >> 1 O 2 u * x * 2 1 O 2 u * y * 2 1 δ * >> 1 O 2 1 Re δ *2 4
Exact solution for a flat plate While from the momentum conservation equation in the y-direction it can be observed that the pressure in the boundary layer assumes the same value assumed in the external flow. O p * p * δ* << O y * x * O u * v * x * δ * u * v * x * + v * v * y * = p * y * + 1 Re 2 v * x * + 2 v * 2 y * 2 O v * v * v * δ* O y * y * 1 O 2 v * x * 2 δ *2 << 1 O 2 v * y * 2 δ * δ * = 1 2 δ * This means that the pressure depends only on the x-direction and that the pressure gradient in the x- momentum conservation equation can be evaluated at the edge of the velocity boundary layer where u(x,y)=u, leading to: ( ) du x ρu dx = dp dx dp dx = 0 p = const for a flat plate Bernoulli equation 5
Exact solution for a flat plate From the order of magnitude analysis applied to the energy conservation equation: O u * T * x * 1 O v * u * T * x * + v * T * y * = 1 Re Pr * ( ) O( y *) δ t O v * T * y * 1 2 T * + 2 T * x * 2 y * 2 O 2 T * x * 2 1 O 2 T * y * 2 1 δ >> 1 *2 t It can then be derived that: O( Re Pr) 1 O( Re) 1 δ *2 δ t *2 Pr δ* * δ t 2 6
Forced convection over a flat plate The resulting boundary layer equation for a flow over a flat plate are then: Mass conservation equation u x + v y = 0 x-momentum conservation equation u u x + v u y = µ ρ 2 u y 2 Energy conservation equation u T x + v T y = k ρc 2 T y 2 Boundary conditions u = v = 0 and T=T s at y = 0 y u u and T T The exact solution is available! These equations have been solved by Blasius by introducing the stream function: ψ( x,y), defined as u = ψ y,v = ψ x 7
Forced convection over a flat plate: the Blasius solution The Blasius solution is obtained on the basis of a similarity variable: η= y Rex x With f(η) to be calculated as unknown. The velocity components are then calculated as: u* = v= u df = =f' u dη 1 νu df f η 2 x dη Once the velocity field is known, shear stress can be calculated τs (x) = µ u d 2f u = µu y y=0 xν dη 2 2 u τs (x) = C fx ρ 2 η=0 C fx = 0.664 Rex 8
Forced convection over a flat plate: the Blasius solution The mean drag coefficient is then given by: Cf = 1 L 1 C fx dx = 1.33 L 0 ReL The momentum and energy conservation equations becomes formally identical when Pr=1. The solution for u* is then identical to the solution for T* T Ts u T* = = =u* T Ts u Nux = 0.332 Rex Pr = 1 ( ) 9
Forced convection over a flat plate: the Blasius solution The analogy between velocity and temperature fields fails when Pr is far from unity. The temperature profiles within the boundary have been calculated by Pohlhausen, who evidenced a strong dependence of profiles from Prandtl number. Pohlhausen proposed that: δ = Pr1 3 Nux = 0.332Re1x 2 Pr1 3, Nu = 0.664 Re1L 2 Pr1 3 δt (0.5 < Pr < 10) 10
Esercizio 1 Applicazioni di Trasmissione del calore, es. 3.1 Una piastra piana, molto larga e lunga 500 mm, è lambita, in situazioni diverse, da diversi fluidi alla pressione atmosferica ed alla temperatura di 20 C; la temperatura della piastra è uniforme e pari a 100 C. Nei casi di aria ad 1,0 m/s, di acqua a 0,10 m/s e di olio minerale leggero ad 1,0 m/s valutare gli spessori degli strati limite della velocità e della temperatura ed i coefficienti locali di trasporto a metà della piastra ed in corrispondenza del bordo di uscita. Determinare, inoltre, i coefficienti medi di trasporto, la resistenza dovuta all attrito e la potenza termica ceduta. ü Numero di Reynolds per stabilire il regime di moto; ü Valutazione dello spessore dello strato limite di velocità e termico; ü Valutazione dei coefficienti locali e medi di scambio termico e di attrito; ü Valutazione della resistenza di attrito e della potenza termica ceduta. Relazioni trovate mediante la soluzione di Pohlhausen per lastra piana. 11
Esercizio 2 Applicazioni di Trasmissione del calore, es. 3.2 Una sottile lastra metallica, quadrata con lato di 17 cm, è lambita parallelamente da aria avente una temperatura di 20 C ed una velocità di 20 m/s. La piastra è isoterma a 100 C. Determinare la potenza termica convettiva dissipata dalla piastra e la potenza meccanica dovuta all attrito. 12